recover-binary-search-tree

Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Note: 
A solution using O(n ) space is pretty straight forward. Could you devise a constant space solution?

confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
   1
  / \
 2   3
    /
   4
    \
     5
The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

public class RecoverSearchTree {
    public static void main(String[] args) {
        TreeNode root =new TreeNode(0);
        root.left = new TreeNode(1);
        RecoverSearchTree test = new RecoverSearchTree();
        test.recoverTree(root);
    }

    private TreeNode firstErrorNode = null;
    private TreeNode secondErrorNode = null;
    private TreeNode preNode = new TreeNode(Integer.MIN_VALUE);
    public void recoverTree(TreeNode root) {
        inOrder(root);
        int temp = firstErrorNode.val;
        firstErrorNode.val = secondErrorNode.val;
        secondErrorNode.val = temp;
    }
    private void inOrder(TreeNode root) {

        if(root == null)
            return;

        inOrder(root.left);

        if(preNode.val >= root.val ){
                if(firstErrorNode == null){
                    firstErrorNode = preNode;
                    secondErrorNode = root;
                }else{
                    secondErrorNode = root;
                }
        }
        //              
        //              ，         preNode  root
        //             root       
        preNode = root;

        inOrder(root.right);

    }
}